Finding container dimensions from a volume constraint

Expert intro: This is a modeling problem where the given relationships among dimensions let us turn a volume equation into a single-variable equation.

Detailed walkthrough

Translate the word problem into equations

Let the width be $w$ meters. Then the length is one meter longer than the width, so

$l=w+1.$

The height is one meter greater than twice the width, so

$h=2w+1.$

The volume of a rectangular solid is

$V=lwh,$

and the problem says the volume must be 100 cubic meters. Substitute the expressions above:

$100=(w+1)w(2w+1).$

Simplify and solve for the width

Expand the factors:

$100=w(2w^2+3w+1).$

So the equation becomes

$2w^3+3w^2+w-100=0.$

This cubic does not factor nicely by inspection, so a numerical method or graphing calculator is appropriate. Testing values near $w=3$ and $w=3.2$ gives

• $w=3$: $3\cdot4\cdot7=84$
• $w=3.2$: $3.2\cdot4.2\cdot7.4\approx99.456$

So the width is a little above 3.2. Solving numerically gives

$w\approx 3.20\text{ m}.$

Find the other dimensions

Now compute the length and height:

$l=w+1\approx 4.20\text{ m},$

$h=2w+1\approx 7.40\text{ m}.$

So the container dimensions are approximately

$3.20\text{ m} \times 4.20\text{ m} \times 7.40\text{ m}.$

Check the result

Multiply the values to verify the volume:

$3.20\cdot 4.20\cdot 7.40\approx 99.456,$

which is very close to 100. Small rounding differences explain the slight mismatch. Using a more precise root gives exactly 100.

Why this setup works

The key algebra step is expressing all dimensions in terms of one variable. Once the relationships are translated correctly, the volume equation becomes a single equation in $w$. This is the standard method for geometric modeling problems with linked dimensions.

💡 Pitfall guide

A frequent mistake is writing the volume equation as $100=(w^2+1)w$ or dropping the middle term when expanding $(w+1)(2w+1)$. The product is $(2w^2+3w+1)$, not $w^2+1$. Another issue is stopping too early after guessing a width near 3.2 without checking the associated length and height. In modeling problems, the relationships among dimensions matter as much as the numerical answer, and a rounded value should always be interpreted as an approximation.

🔄 Real-world variant

If the statement changed to "the length is 2 meters longer than the width" and "the height is 1 meter greater than the width," then the equations would become $l=w+2$ and $h=w+1$. The volume equation would be $100=w(w+2)(w+1)$, which is a different cubic with different dimensions. The solution method is the same: define one variable, rewrite all dimensions, form the volume equation, and solve. Only the algebraic structure changes.

🔍 Related terms

volume equation, rectangular solid, numerical root